Help with Laplace Transformations and 2nd order ODEsby TFM Tags: laplace, odes, order, transformations 

#55
Nov2608, 01:00 PM

P: 774

For 1/(s+1)^3, let's just look at 1/s^3 ; we know that t^2 is 2/s^3 so 1/s^3 will be the inverse of (1/2)(t^2) right?
but, we dont want 1/s^3, we want 1/(s+1)^3, so what do we do (like we did before)? 



#56
Nov2608, 01:13 PM

P: 1,031

First Shft Theorem
[tex] \frac{1}{2}t^2 = \frac{1}{s^3} [/tex] using FST [tex] \frac{1}{2}t^2 = \frac{1}{b^3} [/tex], where b = s + 1 would this make it: \frac{1}{2}(t+1)^2 = \frac{1}{b^3} [/tex] ??? TFM 



#57
Nov2608, 03:54 PM

P: 774

I dont see any e's in your answer (there should be).
now as we did before, the second fraction is the laplace transform of: 4te^(2t) because if you find the laplace of 4t, this will give you 4/s^2 but we dont want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(2t). now, can you do the same for the 1/(s+1)^3 



#58
Nov2708, 08:37 AM

P: 1,031

Okay so:
[tex] t^2 would give: \frac{2}{s^3} [/tex] thus would: [tex] \frac{2}{(s + 1)^3} be \frac{1}{2}t^2 e^{t} [/tex] ??? TFM 



#59
Nov2708, 02:42 PM

P: 774

very good!!




#60
Nov2708, 02:51 PM

P: 1,031

Great
So now: [tex] Y = \frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2} [/tex] this goes to: [tex] Y = \frac{1}{2}t^2 e^{t} + 4te^{2t} [/tex] So now we put this value of Y back into: [tex] s^2(Y)  4 + 4(sY) + 4Y = \frac{1}{s + 2} [/tex] ??? TFM 



#61
Nov2708, 03:02 PM

P: 774

y(t)=(1/2)t^2(e^t) + 4t(e^2t)
This is the answer. dont substitute anything into anything. 



#62
Nov2708, 03:06 PM

P: 1,031

Excellent.
So [tex] y(t) = \frac{1}{2}t^2 e^{t} + 4te^{2t} [/tex] Tomorrow, I'll start (c), if thats okay? TFM 



#63
Nov2808, 03:45 PM

P: 774

yep thats fine




#64
Nov2908, 12:12 PM

P: 1,031

Okay, so:
(c): [tex] y'' + y = sin(t), y_0 = 1, y_0' = 0 [/tex] L(y(t)) = Y L(y'(t)) = sL(y(t))  y(0) L(y''(t)) = sL(y'(t))y'(0) = s^2(L(y(t))sy(0)y'(0) there is no y', [tex] L(y(t)) = Y [/tex] [tex] L(y'(t)) = sL(y(t))  1 [/tex] [tex] L(y''(t)) = sL(y'(t))0 = s^2(L(y(t))sy(0)y'(0) [/tex] and for the sin(t) [tex] sin(\alpha t) = \frac{\alpha}{p^2 + \alpha^2} [/tex] since alpha = 1: [tex] sin(t) = \frac{1}{p^2 + 1} [/tex] Is this okay, I am not quite sure because the lack of y' TFM 



#65
Nov2908, 12:19 PM

P: 774

That is absolutely right.
what next? 



#66
Nov2908, 12:30 PM

P: 1,031

Okay so:
[tex] L(y(t)) = Y [/tex] [tex] L(y'(t)) = sL(y(t))  1 [/tex] [tex] L(y''(t)) = s^2(L(y(t))sy(0) [/tex] So now I substitute these into the original equation: [tex] y'' + y = sin(t) [/tex] [tex] s^2(L(y(t))sy(0) + Y = \frac{1}{p^2 + 1} [/tex] Is this okay? Do I substitute y(0) into sy(0), or is there another value/leave it alone? TFM 



#67
Nov2908, 12:49 PM

P: 774

That's fine
substitute y(0) and also substitue the value L(y(t)) 



#68
Nov2908, 12:59 PM

P: 1,031

Okay, so
[tex] s^2(L(y(t))sy(0) + Y = \frac{1}{p^2 + 1} [/tex] [tex] s^2(Y)1 + Y = \frac{1}{p^2 + 1} [/tex] [tex] Ys^2 + Y  1 = \frac{1}{p^2 + 1} [/tex] Now I need to make the Y the subject: [tex] Ys^2 + Y = \frac{1}{p^2 + 1} + 1 [/tex] factorise out [tex] Y(s^2 + 1) = \frac{1}{p^2 + 1} + 1 [/tex] divide through: [tex] Y = \frac{\frac{1}{p^2 + 1} + 1}{s^2 + 1} [/tex] Now I need to find the inverse. Firstly, split it up into two fractions: [tex] Y = \frac{\frac{1}{p^2 + 1}}{s^2 + 1} + \frac{1}{s^2 + 1} [/tex] Does this look okay? TFM 



#69
Nov2908, 01:42 PM

P: 774

in ur second line, you have: s^y1+y=1/(s^2+1)
the 1 on the left hand side should be s (since you have:L(y'')=s(sy1) and remember: s is p so dont put both...choose one of them :) 



#70
Nov2908, 01:52 PM

P: 1,031

Okay so:
[tex] s^2(Y)1 + Y = \frac{1}{p^2 + 1} [/tex] [tex] Ys^2 + sY  1 = \frac{1}{s^2 + 1} [/tex] Does this look okay now? TFM 



#71
Nov2908, 03:35 PM

P: 774

you should have:
Ys^2  s + Y=1/(s^2+1) i think you made a mistake while substituting. 



#72
Nov2908, 03:58 PM

P: 1,031

Okay, back a bit originally:
[tex] s^2(L(y(t))sy(0) + Y = \frac{1}{s^2 + 1} [/tex] y(0) = 1 L(y(t)) = Y put these in: [tex] s^2(Y)  1 + Y = \frac{1}{s^2 + 1} [/tex] goes to: [tex] Ys^2  1 + Y = \frac{1}{s^2 + 1} [/tex] I still have a minus one where there should be a minus s.???? TFM 


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